Question

help explain this



http://prntscr.com/1at9sj

Answer 1

what do you think? any thoughts ?

Answer 2

yeah i think it is a circle
option B

Answer 3

yes it is a circle, correct.
From the given equation, can you tell me the co-ordinates of the center of the given circle ?

Answer 4

i have to solve this equation and turn it into this form
(x-h)2+(y-k)2=r2
and then the center will be (h,k)
i can't tell you directly, can you?

Answer 5

Whatever you said above was correct.
(h,k) is the center of (x-h)^2 + (y-k)^2 = r^2

now when you expand all the terms,

x^2 -2xh + h^2 + y^2 - 2yk + k^2 = r^2

x^2 + y^2 -2hx -2ky + h^2 + k^2 - r^2 =0

Your given equation is of this form only.
Now recall that (h,k) was the center
that means if the expression is represented like this, then negative halves of the coefficients of x and y are the respective co-ordinates of the center of the circle.

for example,
in x^2 + y^2 - 2x + 4y + 2 = 0

(-(-2)/2 , -(4)/2 ) are the co-ordinates of the center
i.e. (1,-2) is the center
now, using this, can you tell the center of your given equation ?

Answer 6

oh yeah, this is easy, this way i don't have to solve the entire equation!
the center is (2,2)

Answer 7

thats right.

Answer 8

btw how do you distinguish a circle equation from a ellipse equation?

Answer 9

in circle, coefficients of x^2 and y^2 = 1
not in ellipse

Answer 10

here is another thing then, not related to your question much tough :
in x^2 + y^2 - 2gx - 2fy + c =0
you already know center is (g,f)
now for the radius, recall the above expression and compare and see that
c= h^2 + k^2 - r^2

so r =sqrt( h^2 + k^2 - c)

you should probably memorize this stuff in case you are preparing for some competitive exam .

Answer 11

the above equation is of ellipse right?

Answer 12

what above equation ?

Answer 13

x^2 + y^2 - 2gx - 2fy + c =0 this

Answer 14

here coefficients of x^2 and y^2 both =1
hence its a cricle
had it not been 1, then ellipse

Answer 15

oh yeah yeah i got this,
you explained it superbly thanks @shubhamsrg :D

Answer 16

glad to help (Y)